[ION 2 \( 7 ; x ; y ;-11 ; \ldots \) is an arithmetic sequence. Determine the values of \( x \) and \( y \). (4) Given the quadratic number pattern: \( -3 ; 6 ; 27 ; 60 ; \ldots \) 2.2.1 Determine the general term of the pattern in the form \( T_{n}=a n^{2}+b n+c \). 2.2.2 Calculate the value of the \( 50^{\text {th }} \) term of the pattern. 2.2.3 Show that the sum of the first \( n \) first-differences of this pattern can be given by \( S_{n}=6 n^{2}+3 n \). 2.2.4 How many consecutive first-differences were added to the first term of the quadratic number pattern to obtain a term in the quadratic number pattern that has a value of 21060 ? [17 ION 3 Prove that \( \sum_{k=1}^{\infty} 4.3^{2-k} \) is a convergent geometric series. Show ALL your calculations. If \( \sum_{k=p}^{\infty} 4.3^{2-k}=\frac{2}{9} \), determine the value of \( p \). (3

Sure, let's tackle each of these problems step by step. --- ## **ION 2** ### **Problem 7** **Given:** An arithmetic sequence \( 7 ; x ; y ; -11 ; \ldots \) **Find:** The values of \( x \) and \( y \). **Solution:** In an arithmetic sequence, the difference between consecutive terms is constant. Let's denote this common difference as \( d \). Given the sequence: - First term (\( a_1 \)) = 7 - Second term (\( a_2 \)) = \( x \) - Third term (\( a_3 \)) = \( y \) - Fourth term (\( a_4 \)) = -11 Using the property of arithmetic sequences: \[ a_{n} = a_{1} + (n-1)d \] Let's find the common difference \( d \) using the first and fourth terms. \[ a_4 = a_1 + 3d \\ -11 = 7 + 3d \\ -11 - 7 = 3d \\ -18 = 3d \\ d = -6 \] Now, find \( x \) and \( y \): \[ a_2 = a_1 + d = 7 + (-6) = 1 \\ a_3 = a_2 + d = 1 + (-6) = -5 \] **Answer:** - \( x = 1 \) - \( y = -5 \) --- ## **Quadratic Number Pattern** **Given Pattern:** \( -3 ; 6 ; 27 ; 60 ; \ldots \) ### **2.2.1 Determine the general term \( T_{n} = a n^{2} + b n + c \).** **Solution:** Assume the general term is quadratic: \[ T_n = an^2 + bn + c \] We can set up equations using the given terms: 1. For \( n = 1 \): \[ T_1 = a(1)^2 + b(1) + c = a + b + c = -3 \quad \text{(Equation 1)} \] 2. For \( n = 2 \): \[ T_2 = a(2)^2 + b(2) + c = 4a + 2b + c = 6 \quad \text{(Equation 2)} \] 3. For \( n = 3 \): \[ T_3 = a(3)^2 + b(3) + c = 9a + 3b + c = 27 \quad \text{(Equation 3)} \] Now, solve the system of equations. **Subtract Equation 1 from Equation 2:** \[ (4a + 2b + c) - (a + b + c) = 6 - (-3) \\ 3a + b = 9 \quad \text{(Equation 4)} \] **Subtract Equation 2 from Equation 3:** \[ (9a + 3b + c) - (4a + 2b + c) = 27 - 6 \\ 5a + b = 21 \quad \text{(Equation 5)} \] **Subtract Equation 4 from Equation 5:** \[ (5a + b) - (3a + b) = 21 - 9 \\ 2a = 12 \\ a = 6 \] **Substitute \( a = 6 \) into Equation 4:** \[ 3(6) + b = 9 \\ 18 + b = 9 \\ b = -9 \] **Substitute \( a = 6 \) and \( b = -9 \) into Equation 1:** \[ 6 - 9 + c = -3 \\ -3 + c = -3 \\ c = 0 \] **Answer:** \[ T_n = 6n^2 - 9n \] --- ### **2.2.2 Calculate the value of the \( 50^{\text{th}} \) term of the pattern.** **Solution:** Using the general term \( T_n = 6n^2 - 9n \): \[ T_{50} = 6(50)^2 - 9(50) = 6(2500) - 450 = 15000 - 450 = 14550 \] **Answer:** \( T_{50} = 14,\!550 \) --- ### **2.2.3 Show that the sum of the first \( n \) first-differences of this pattern can be given by \( S_{n} = 6n^{2} + 3n \).** **Solution:** First, find the first difference sequence. The first difference \( D_n = T_n - T_{n-1} \). Given \( T_n = 6n^2 - 9n \): \[ D_n = T_n - T_{n-1} = [6n^2 - 9n] - [6(n-1)^2 - 9(n-1)] \] \[ D_n = 6n^2 - 9n - [6(n^2 - 2n + 1) - 9n + 9] \] \[ D_n = 6n^2 - 9n - 6n^2 + 12n - 6 + 9n - 9 \] \[ D_n = (6n^2 - 6n^2) + (-9n + 12n + 9n) + (-6 - 9) \] \[ D_n = 12n - 15 \] Now, find the sum of the first \( n \) first-differences: \[ S_n = \sum_{k=1}^{n} D_k = \sum_{k=1}^{n} (12k - 15) \] \[ S_n = 12 \sum_{k=1}^{n} k - 15n \] \[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \] \[ S_n = 12 \left( \frac{n(n+1)}{2} \right) - 15n = 6n(n+1) - 15n = 6n^2 + 6n - 15n = 6n^2 - 9n \] This contradicts the desired \( S_n = 6n^2 + 3n \). It seems there might be an error in the initial calculation. Wait, let's re-examine the first difference calculation. **Recalculating \( D_n \):** Given \( T_n = 6n^2 - 9n \), then: \[ D_n = T_n - T_{n-1} = (6n^2 - 9n) - [6(n-1)^2 - 9(n-1)] \] \[ = 6n^2 - 9n - [6(n^2 - 2n + 1) - 9n + 9] \] \[ = 6n^2 - 9n - 6n^2 + 12n - 6 + 9n - 9 \] \[ = (6n^2 - 6n^2) + (-9n + 12n + 9n) + (-6 -9) \] \[ = 12n - 15 \] Thus, \( D_n = 12n - 15 \). Now, sum \( D_n \) from \( k=1 \) to \( n \): \[ S_n = \sum_{k=1}^{n} (12k - 15) = 12 \sum_{k=1}^{n} k - 15n = 12 \cdot \frac{n(n+1)}{2} - 15n = 6n(n+1) - 15n = 6n^2 + 6n - 15n = 6n^2 - 9n \] But the problem states \( S_{n} = 6n^{2} + 3n \). There seems to be a discrepancy. However, considering the pattern provided: Given the sequence: \( -3 ; 6 ; 27 ; 60 ; \ldots \) Compute first differences: \[ 6 - (-3) = 9 \\ 27 - 6 = 21 \\ 60 - 27 = 33 \\ \] Differences: 9, 21, 33,... These differences themselves form an arithmetic sequence with a common difference of 12. Thus, the sum of the first \( n \) differences: \[ S_n = \sum_{k=1}^{n} (9 + 12(k-1)) = \sum_{k=1}^{n} (12k - 3) = 12 \cdot \frac{n(n+1)}{2} - 3n = 6n(n + 1) - 3n = 6n^2 + 6n - 3n = 6n^2 + 3n \] **Answer:** \[ S_n = 6n^2 + 3n \] --- ### **2.2.4 How many consecutive first-differences were added to the first term of the quadratic number pattern to obtain a term in the quadratic number pattern that has a value of 21,060?** **Solution:** We need to find \( n \) such that when adding the first \( n \) first-differences to the first term, the resulting term is 21,060. Given: - First term (\( T_1 \)) = -3 - Sum of first \( n \) first-differences (\( S_n \)) = 6n^2 + 3n - Desired term (\( T \)) = 21,060 The relationship is: \[ T = T_1 + S_n \\ 21,060 = -3 + 6n^2 + 3n \\ 6n^2 + 3n - 21,063 = 0 \\ \] Divide the entire equation by 3 to simplify: \[ 2n^2 + n - 7,021 = 0 \] Now, solve the quadratic equation \( 2n^2 + n - 7021 = 0 \). Use the quadratic formula: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where: - \( a = 2 \) - \( b = 1 \) - \( c = -7,021 \) Calculate the discriminant (\( \Delta \)): \[ \Delta = 1^2 - 4(2)(-7021) = 1 + 56,168 = 56,169 \] \[ \sqrt{56,169} = 237 \] Now, find \( n \): \[ n = \frac{-1 \pm 237}{4} \] We discard the negative solution as \( n \) must be positive: \[ n = \frac{-1 + 237}{4} = \frac{236}{4} = 59 \] **Answer:** 59 consecutive first-differences were added. --- ## **ION 3** ### **Problem** **Given:** \( \sum_{k=1}^{\infty} 4 \cdot 3^{2 - k} \) 1. **Prove that it is a convergent geometric series. Show all your calculations.** 2. **If \( \sum_{k=p}^{\infty} 4 \cdot 3^{2 - k} = \frac{2}{9} \), determine the value of \( p \).** ### **Solution:** #### **1. Prove that it is a convergent geometric series.** **Identify the series:** \[ \sum_{k=1}^{\infty} 4 \cdot 3^{2 - k} \] Simplify the general term: \[ a_k = 4 \cdot 3^{2 - k} = 4 \cdot 3^2 \cdot 3^{-k} = 4 \cdot 9 \cdot \left(\frac{1}{3}\right)^k = 36 \cdot \left(\frac{1}{3}\right)^k \] Recognize the geometric series form: \[ a_k = ar^{k} \] Comparing: \[ ar^{k} = 36 \cdot \left(\frac{1}{3}\right)^k \] Thus, the first term when \( k = 1 \): \[ a = 36 \cdot \left(\frac{1}{3}\right)^1 = 36 \cdot \frac{1}{3} = 12 \] The common ratio is: \[ r = \frac{1}{3} \] **Criteria for convergence of a geometric series:** A geometric series \( \sum ar^{k} \) converges if \( |r| < 1 \). Here, \( r = \frac{1}{3} \), and \( \left| \frac{1}{3} \right| < 1 \). **Conclusion:** Since \( |r| = \frac{1}{3} < 1 \), the series is convergent. **Answer:** The series is a convergent geometric series because the common ratio \( r = \frac{1}{3} \) satisfies \( |r| < 1 \). #### **2. Determine the value of \( p \) if \( \sum_{k=p}^{\infty} 4 \cdot 3^{2 - k} = \frac{2}{9} \).** **Solution:** Given: \[ \sum_{k=p}^{\infty} 4 \cdot 3^{2 - k} = \frac{2}{9} \] Simplify the series: \[ \sum_{k=p}^{\infty} 4 \cdot 3^{2 - k} = \sum_{k=p}^{\infty} 36 \cdot \left(\frac{1}{3}\right)^k \] This is a geometric series with: \[ a = 36 \cdot \left(\frac{1}{3}\right)^p \quad \text{and} \quad r = \frac{1}{3} \] The sum of an infinite geometric series starting from \( k = p \): \[ S = \frac{a}{1 - r} = \frac{36 \cdot \left(\frac{1}{3}\right)^p}{1 - \frac{1}{3}} = \frac{36 \cdot \left(\frac{1}{3}\right)^p}{\frac{2}{3}} = 36 \cdot \left(\frac{1}{3}\right)^p \cdot \frac{3}{2} = 54 \cdot \left(\frac{1}{3}\right)^p \] Set this equal to \( \frac{2}{9} \): \[ 54 \cdot \left(\frac{1}{3}\right)^p = \frac{2}{9} \] Divide both sides by 54: \[ \left(\frac{1}{3}\right)^p = \frac{2}{9} \div 54 = \frac{2}{9 \times 54} = \frac{2}{486} = \frac{1}{243} \] Express \( \frac{1}{243} \) as \( \left(\frac{1}{3}\right)^5 \) since \( 3^5 = 243 \): \[ \left(\frac{1}{3}\right)^p = \left(\frac{1}{3}\right)^5 \] Therefore: \[ p = 5 \] **Answer:** \( p = 5 \) ---